Impulse Momentum Method for Rotational

[freshbox]

Homework Statement

I have abit of problem understanding this 2 question, I would appreciate if someone can guide me along.

For diagram 1 (Left)
I interpret the condition as: At 0sec, initial velocity is 0, At 5sec final velocity is unknown.


For diagram 2 (Right)
I interpret the condition as: At 0 sec, initial velocity is 0, At 3sec final velocity is 14.5m/s


Please let me know if I get any of the condition wrong, thank you.

 

[voko]

I think you understand the conditions correctly.

 

[freshbox]

o_0 hmm strange then how come i cannot get the answer...

 

[voko]

What goes wrong?

 

[freshbox]

For Diagram 1 part A

M₁=mv
=0

M₂=mv
=(20)(V₂)
=20V₂

M₁+I_1-2=M₂
981-5TA=20V₂
5TA=981-20V₂
TA=196.2-4V₂

Ans is TA=196.2-4V

4V₂ and 4V are both different meaning right?

 

[voko]

Why are you ignoring the tangential force?

 

[freshbox]

You mean the 75N force?

What has it got to do with part A?

 

[voko]

You have already solved part A as far as I can tell.

 

[freshbox]

My Answer is: TA=196.2-4V₂
Question Ans is : TA=196.2-4V

Is there any difference between V₂ and V? one is initial one is final.

 

[voko]

The initial velocity is zero per the conditions ("released at t = 0"). So v can only be the final.

 

[freshbox]

How about the subscript? Don't have to write?

 

[voko]

Up to you. You can say "let's denote the velocity at t = 5 s as V" or in any other way. That does not change the underlying physics. You just need to know what denotes what.

 

[freshbox]

Part B

M₁=Iω
=0

M₂=Iω
=(2.5)(V/0.15)
=16.67V

I_1-2=Torque x Time
=[-22.5+Ta(0.15)]5
=34.65-3V

M₁+I_1-2=M₂
34.65-3V=16.67V
34.65=19.67V
V=1.76

V=rω
ω=1.76/0.15
ω=11.74 - Ans Wrong

 

[voko]

ω=11.74 - Ans Wrong

Hmm. I have used a somewhat different method, but got the same result. Let's see if somebody else here can point out our mistake.

 

[Doc Al]

M₂=Iω
=(2.5)(V/0.15)
=16.67V

OK.

I_1-2=Torque x Time
=[-22.5+Ta(0.15)]5
=34.65-3V

Where did you get the 22.5 from? Redo that calculation.

 

[freshbox]

Hi Doc Al, thanks for helping.

I got my -22.5 from the tangential force Ft=75N I used the 75 times 0.3 = 22.5 Because Tension x Radius = Torque


*Ok i think i got it, it's 300mm, 150mm diameters

 

[Doc Al]

*Ok i think i got it, it's 300mm, 150mm diameters

There you go.

 

[freshbox]

M₁=Iω
=0

M₂=Iω
=(2.5)(V/0.15)
=16.67V

I_1-2= Torque x Time
=(TaX0.075-75X0.15)5
=17.325-1.5V

M₁+I_1-2=M₂
17.325-1.5v=16.67v
17.325=18.17v
v=0.95

v=rω
0.95=0.15ω
ω=6.3 wrong again

 

[Doc Al]

M₂=Iω
=(2.5)(V/0.15)
=16.67V

Same problem with radius vs diameter. (Sorry I didn't spot it before.)

 

[freshbox]

how come i need to take the radius of the axle and not the wheel?

 

[Doc Al]

how come i need to take the radius of the axle and not the wheel?

You are relating ω to the speed V, which is the tangential speed of the axle edge, not the wheel edge.

 

[freshbox]

Can I say the "v" that I am finding is actually the velocity of the axle. After getting the velocity of the axle, I divide by 0.075 (axle radius) is because in a compound pulley, Angular velocity are the same. I am just using the axle velocity to get ω.

 

[Doc Al]

Can I say the "v" that I am finding is actually the velocity of the axle. After getting the velocity of the axle, I divide by 0.075 (axle radius) is because in a compound pulley, Angular velocity are the same. I am just using the axle velocity to get ω.

Yes. V is the velocity of the falling mass, and since its cord is wrapped around the axle, V is also the tangential velocity of the axle.

The axle and wheel are attached, so they have a common ω.

 

[freshbox]

Ok..I'm sorry can you help me take a look at this question part C

M₁=Iω
=(3.5)(12.5)
=43.75

M₂
=Iω
=3.5ω

I_1-2=Torque X Time
=[(-TA)(0.2)+(TB)(0.5)]5
=-32ω+3783.5

M₁+I_1-2=M₂
43.75-32ω+3783.5=3.5ω
3827.25=35.5ω
ω=107.80 - Wrong Answer

 

[Doc Al]

M₁=Iω
=(3.5)(12.5)
=43.75

M₂
=Iω
=3.5ω

I_1-2=Torque X Time
=[(-TA)(0.2)+(TB)(0.5)]5
=-32ω+3783.5

Double check that last calculation.

 

[freshbox]

=[(-TA)(0.2)+(TB)(0.5)]5

=[(-2ω+465.5)0.2+(1327.2-12ω)0.5]5

=(-0.4ω+93.1+663.6-6ω)5

=(-6.4ω+756.7)5

=-32ω+3783.5

I'm abit blind hehe can you tell me where is wrong?

 

[Doc Al]

=[(-TA)(0.2)+(TB)(0.5)]5

=[(-2ω+465.5)0.2+(1327.2-12ω)0.5]5

Looks to me like you left out a minus sign.

 

[freshbox]

fail -.- what a mess...hehe anyway thanks for your time and help (voko too), thanks :)

 

[voko]

I can't believe I mistook the diameters for the radii, but I did. Not just once, but multiple times when I cross-checked my results via at least three different methods. Oh well. Thanks Doc Al for pointing that out.